Question: Let $x$ be a positive integer such that $9x\equiv 1\pmod{25}$.

What is the remainder when $11+x$ is divided by $25$?
Solution: The given information can be expressed by writing $x\equiv 9^{-1}\pmod{25}$. Thus we wish to compute $11+9^{-1}\pmod{25}$.

Modulo $25$, we can write $11$ as $11\cdot (9\cdot 9^{-1}) \equiv (11\cdot 9)\cdot 9^{-1} \equiv 99\cdot 9^{-1}$. Thus \begin{align*}
11 + 9^{-1} &\equiv 99\cdot 9^{-1} + 1\cdot 9^{-1} \\
&\equiv 100\cdot 9^{-1} \\
&\equiv 0\cdot 9^{-1} \\
&\equiv 0\pmod{25},
\end{align*}so the remainder when $11+x$ is divided by $25$ is $\boxed{0}$.

Notice that the trick we used here is analogous to using a common denominator to add fractions.